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 Maths question, now with free magic trick!

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Roycropper
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PostPosted: Wed Mar 19, 2008 7:45 pm Reply with quoteBack to top

Jim is at the Western Union office, sending $293 to Usman for his barrister fee. He makes a mistake and sends $239 by accident.

This is called a transposition error. They teach you at accountancy school that if you have a difference after adding up numbers, try dividing the difference by 9. If it goes exactly you may have transposed 2 numbers. In this case 293-239=54. 54/9=6. This always works. The first number it works for is 12 (you cant transpose 11) 21-12=9

I was explaining this to Hayley last night as it was part of the answer to a much more complicated puzzle. My question is WHY? I know it does work, but there must be an explanation.

Answer me this and I'll show you a good magic trick that uses this principle.

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Last edited by Roycropper on Thu Mar 20, 2008 9:56 am; edited 1 time in total
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Chibuike
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PostPosted: Wed Mar 19, 2008 7:49 pm Reply with quoteBack to top

er um, I wan't good in math. Thank God for calculators!

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PostPosted: Wed Mar 19, 2008 7:54 pm Reply with quoteBack to top

more info here *** clicky ***

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PostPosted: Wed Mar 19, 2008 8:02 pm Reply with quoteBack to top

Here you go Roy:

Any permutation can be obtained by a sequence of interchanges in
which we swap one pair of digits. So it is enough to show that the
difference is divisible by 9 if we swap any two digits. (You can fill
in the details.)

Suppose the digits of the number are a_n ... a_1 a_0, so that the
value of the number is

a_n*10^n + ... + a_i*10^i + ... + a_j*10^j + ... + a_1*10 + a_0

Then if we swap digits a_i and a_j (I'll assume i>j), the new value
will be

a_n*10^n + ... + a_j*10^i + ... + a_i*10^j + ... + a_1*10 + a_0

The difference is then

(a_i*10^i + a_j*10^j) - (a_j*10^i + a_i*10^j)

= (a_i - a_j)*10^i - (a_i - a_j)*10^j

= (a_i - a_j)(10^i - 10^j)

= (a_i - a_j)(10^(i-j) - 1)10^j

So we can prove our goal if we can show that

10^n - 1 = 9k

for some k, for any positive integer n. (Think about writing such a
number out, and you will immediately see that it is true; but we
still need to prove it.)

Now consider the polynomial

x^n - 1

Since it is zero when x=1, it has (x-1) as a factor. For example,
x^2 - 1 = (x+1)(x-1).

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PostPosted: Wed Mar 19, 2008 8:02 pm Reply with quoteBack to top

simple:
first digit=a
second digit=b

You claim (10*a+b) - (10*b+a) = x times 9

10a+b-10b-a=9a-9b so what's there to understand Very Happy

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PostPosted: Wed Mar 19, 2008 8:05 pm Reply with quoteBack to top

(scratching head) huh?

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PostPosted: Wed Mar 19, 2008 8:19 pm Reply with quoteBack to top

Okay in numbers then pick two for me. 2 and 7 ? Thanks Very Happy

he says 72 - 27 is a natural number times 9, so I write

(70 + 2) -(20+7) = 45 or 5 times 9 so he is right, now when I fill it in with letters you can see that it is alway a 9 no matter what the digits are.

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PostPosted: Wed Mar 19, 2008 8:32 pm Reply with quoteBack to top

ahhhh I think I got it

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PostPosted: Wed Mar 19, 2008 8:50 pm Reply with quoteBack to top

Good Smile and budanzig wrote down the same a bit different, but I'm lazy Very Happy

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Roycropper
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PostPosted: Thu Mar 20, 2008 9:54 am Reply with quoteBack to top

Ok cheers, now I can explain it to Hayley. Laughing

TRICK:

Take your phone book, find page 108, count 9 names down, and write that name and number down. seal it in an envelope

Ask someone to pick a 3 digit number, or 3 people to pick a digit each. Write that number down. Then reverse it, and write that down. so for example we have:

459 and 954. take the smaller number from the bigger one.

954
459 -
495

Reverse that number, ie 495 becomes 594, then add those 2 numbers, The result is always 1089. Ask someone to turn to page 108 of the phone book, and count down 9 places. The name and number will be the same as in your envelope! Shocked

If the answer isn't 1089, you made a mistake.

As far as I can get my head round it, you have only transposed the first and last digit in stage 1, (the middle one stays where it is) so the number 9 rears its head, no doubt due to the earlier proof. The difference divides by 9, in this case 495/9=55. It also divides by 11, as far as I can see.

When you add the 495 to 594, look what happens:

495
594

You always get 2 9's on the middle, flanked by one either side. So the 9's are in the ratio 1:2:1. 121 x 9 = 1089. QED, kindof. (121 also = 11 x 11) - confusing myself now, but it's true.

Now try this on your mugu. If you can get your hands on a Lagos phone directory, all the better.

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Last edited by Roycropper on Thu Mar 20, 2008 9:51 pm; edited 1 time in total
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Breddan Butter
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PostPosted: Thu Mar 20, 2008 10:54 am Reply with quoteBack to top

Roy,
You should explain that, when you ask someone to pick a number, the first and last digits must differ by at least 2, otherwise the trick won't work.

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thud419
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PostPosted: Thu Mar 20, 2008 12:08 pm Reply with quoteBack to top

^^^ I disagree. The sole condition is that the first and third digits are different, which can be assured by asking the audience for three different digits, although the value of the middle one does not matter.

OK, Here's how it works:

You have a number abc. let's assume a > c, because otherwise you would do the subtraction the other way round, which confuses things without changing the argument.

abc - cba = def

From our previous proof, we know that def is divisible by nine. However we could have done:

ac - ca = gh, and gh would also have been divisible by nine.

We know that a > c, and therefore there must be a borrow in from the tens column, but b=b, which means that there will also be a borrow in from the hundreds column.

So in both cases the borrows are the same, and therefore g=d and h=f.

Now we have def is divisible by nine and df is divisible by nine. It follows that e must be divisible by nine, and we also find that it is a result of a zero subtraction minus a borrow of one. It must be 9.

d is the result of the subtraction of two single digits, less a borrow of 1. It can never be 9.

df is divisible by nine and therefore d+f is a multiple of nine. The only values of df that can cause us difficulties are 99 and 0. Neither can arise given that a > c and d < 9.

Our second stage is def + fed = 1089

We know d+f = 9 and e = 9, giving the 121*9 result analysed above.

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Breddan Butter
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PostPosted: Thu Mar 20, 2008 2:03 pm Reply with quoteBack to top

That's all very impressive. Smile

I tried Roy's 'trick' with the initial chosen number of 413, and then again with 514.
It doesn't compute to 1089. Crying or Very sad

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seton
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PostPosted: Thu Mar 20, 2008 2:32 pm Reply with quoteBack to top

^^ I guess you know this, but what you should be doing is 099 + 990 Smile

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PostPosted: Thu Mar 20, 2008 2:33 pm Reply with quoteBack to top

Technically it does as long as you maintain 3 digits.

413
-314
_____
099
+990
_____
1089

Its one of the oldest mentalism tricks in the book. I always use a "prediction" on a piece of paper and fold it up and clip it underneath the clip on the pen so I cannot "change" it. Then have them use the same pen to do the work.
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Breddan Butter
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PostPosted: Thu Mar 20, 2008 2:41 pm Reply with quoteBack to top

seton wrote:
^^ I guess you know this, . . . .

Guess again; I didn't know that, and it sure wasn't obvious to me. Smile

Thanks for the enlightenment.

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PostPosted: Thu Mar 20, 2008 2:54 pm Reply with quoteBack to top

depends how you do it Smile

number=413 reverse=314 diff.=099 reverse of diff.= 990 then it works. Wink
Those fl**ing mathematicians Very Happy
You should have a new restriction not a>c but a > (c+1) and then it works. And of course c > 0 Laughing

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PostPosted: Thu Mar 20, 2008 3:12 pm Reply with quoteBack to top

Hm....
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PostPosted: Thu Mar 20, 2008 8:29 pm Reply with quoteBack to top

Image
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